Question

Match the entries of Column I with those of Column II.
The lines given by left hand side in Column I have the property mentioned in Column II,

Answer 1

$\left(a\right)\to \left(q\right)$ as $a+b=0$. Its intersection with x-axis $(y=0)$ is given by
$6x^2-x-1=0$ or $(3x+1)(2x-1)=0$ $\therefore x=\frac{1}{2},-\frac{1}{3}$
$\therefore I_x=x_1-x_2=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ $\therefore \left(a\right)\to \left(r\right)$
Its interaction with y-axis $(x=0)$ is given by
$-6y^2+5Y-1=0$ or $6y^2-5y+1=0$
or $(3y-1)(2y-1)=0$ $\therefore y=\frac{1}{2},\frac{1}{3}$.
$\therefore I_y=y_1-y_2=\frac{1}{2}-\frac{1}{3}=\frac{5}{6}$ $\therefore \left(a\right)\to \left(s\right)$
$\therefore \left(a\right)\to (q,r,s)$
$\left(b\right)\to (r,s),\left(c\right)\to (r,s)$ Proceed as in $\left(a\right)$
(d) $6(x+y{)}^2-7(x+y)+1=0$ or $6t^2-7t+1=0$
$\therefore (t-1)(6t-1)=0$
$\therefore x+y-1=0$ or $x+y-\frac{1}{6}=0$
$\therefore$ Parallel lines $\left(d\right)\to \left(p\right)$
$y=0,x=1,\frac{1}{6}$ $\therefore I_x=1-\frac{1}{6}=\frac{5}{6}$ $\therefore \left(d\right)\to \left(r\right)$
$x=0,y=1,\frac{1}{6}$ $\therefore I_y=1-\frac{1}{6}=\frac{5}{6}$ $\therefore \left(d\right)\to (p,r)$